Subject: Re: [xsl] Can you pass a parameter from a stylesheet to the resulting HTML file? From: Joerg Heinicke <joerg.heinicke@xxxxxx> Date: Tue, 04 Jun 2002 20:55:30 +0200 |
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xhtml="http://www.w3.org/1999/xhtml" version="1.0">
<xsl:output method="xml" encoding="UTF-8"/>
<xsl:template match="node() | @*"> <xsl:copy> <xsl:apply-templates select="node() | @*"/> </xsl:copy> </xsl:template>
<xsl:template match="xhtml:ol[@start='var']"> <xsl:copy> <xsl:apply-templates select="@*"/> <xsl:attribute name="start"> <xsl:value-of select="count(preceding::xhtml:ol/xhtml:li) + 1"/> </xsl:attribute> <xsl:apply-templates select="node()"/> </xsl:copy> </xsl:template>
</xsl:stylesheet>
The second xsl file filters out any <li> item with a class of "mgronly":
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xhtml="http://www.w3.org/1999/xhtml" version="1.0">
<xsl:output method="xml" encoding="UTF-8"/>
<xsl:template match="node() | @*"> <xsl:copy> <xsl:apply-templates select="node() | @*"/> </xsl:copy> </xsl:template>
<xsl:template match="xhtml:li[@class='mgronly']"/>
</xsl:stylesheet>
Here's what I'm trying to do: In the first transformation, I need to replace "var" in the second <ol> tag with "4". In the second transformation, because one of the preceding <li> items is filtered out, I need to replace "var" with "3". (I don't need to have the XSL files count the <li> items to compute the "4" and "3", unless it's very easy. I can just hard code "4" and "3".) Is there a way to do this?
Thanks again for all your help--
Kathryn
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